LightOJ 1341

http://www.lightoj.com/volume_showproblem.php?problem=1341

题意：给你长方形的面积a，边最小为b，问有几种情况。

思路：对a进行素因子分解，再乘法原理算一下，最后减去小于b的因子的情况即可。

/** @Date    : 2016-12-01-19.04
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version :
  */

#include<bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e6+20;


LL pri[N];
int c = 0;
bool vis[N];

void prime()
{
    for(int i = 2; i <= N; i++)
    {
        if(!vis[i])
        {
            for(int j = i + i; j <= N; j+= i)
            {
                if(!vis[j])
                    vis[j] = 1;
            }
            pri[c++] = i;
        }
    }
}
int main()
{
    prime();
    int T;
    int cnt = 0;
    cin >> T;

    while(T--)
    {
        LL a , b;
        scanf("%lld%lld", &a, &b);

        LL t = a;
        LL ans = 1;
        LL cc = 0;
        if(b > sqrt(a))
        {
            printf("Case %d: 0
", ++cnt);
            continue;
        }
        for(int i = 0; i < c && pri[i]*pri[i] <= t; i++)
        {
            LL cc = 0;
            while(t % pri[i] == 0)
            {
                cc++;
                t /= pri[i];
            }
            ans *= cc + 1;
        }
        if(t > 1)
            ans *= 2;

        ans /= 2;
        for(int i = 1; i < b; i++)
            if(a % i == 0)
            {
                ans--;
            }

        printf("Case %d: %lld
", ++cnt, ans);
    }
    return 0;
}